The max–min newsvendor pricing problem under conditional value-at-risk criterion

Abstract

This paper studies a risk-averse newsvendor pricing model with limited demand information under the conditional value-at-risk (CVaR) criterion. The paper uses a max–min approach and the objective is to maximize the lower bound on the CVaR of the loss, i.e., the negative of profit in the worst possible distribution case. The paper analyzes the optimal ordering and pricing decisions under both multiplicative and additive demand models, identifies the optimality conditions of the lower bound on the CVaR of loss, and obtains the implicit solutions for the optimal price and order quantity. Furthermore, the paper analyzes the sensitivity of optimal solutions with respect to the degree of risk aversion.

Appendices

Appendix A: Derivations

Proof

1. The detailed derivation of the two-point cumulative distribution in Eqs. (15)–(18). We will show that there exists a two-point distribution $D^{*}\thicksim \Phi ^{*}\in \mathscr {F}$ such that $E\vert q-D\vert \le E\vert q-D^{*}\vert$ for all $D\thicksim \Phi \in \mathscr {F}$.

First, the following proof will use the known fact that for any quadratic polynomial function H(x), we have $E[H(D_1)]=E[H(D_2)]$ for all $D_1,D_2\in \mathscr {F}$ (Scarf 1958). That is, if two random variables have the same expectation and the same variance, then their quadratic polynomial functions have the same expectation.

Second, we construct a quadratic polynomial function $H_{0}(x)$, which satisfies $H_{0}(x)\ge \vert q-x\vert$ for all $x\ge 0$ and $H_{0}(x)$ is tangent to $\vert q-x\vert$ at two points $x_1,x_2$, i.e., $H_{0}(x_1)=\vert q-x_1\vert$ and $H_{0}(x_2)=\vert q-x_2\vert$. It is easy to know that $x_1,x_2$ are symmetric with respect to $x=q$, i.e., there exists e such that $x_1=q-e$ and $x_2=q+e$.

Finally, define a two-point distribution $D^{*}$, which assigns weight $\beta$ to $q-e$ and weight $1-\beta$ to $q+e$. It is easy to verify that $e=(v^2+(q-z)^2)^{1/2}$ and $\beta =\frac{(v^2+(q-z)^2)^{1/2}+(q-z)}{2(v^2+(q-z)^2)^{1/2}}$ to ensure $E[D]=z,Var[D]=v^2$, i.e, $D^{*}\thicksim \Phi ^{*}\in \mathscr {F}$.

Based on the above facts, we have $E\vert q-D\vert \le E[H_{0}(D)]=E[H_{0}(D^{*})]=E\vert q-D^{*}\vert$. The first inequality is based on $H_{0}(x)\ge \vert q-x\vert$ for all $x\ge 0$, the second equality is based on $D,D^{*}\in \mathscr {F}$, and the third equality is based on that $D^{*}$ is a two-point distribution at $x_1,x_2$ and $H_{0}(x)$ is tangent to $\vert q-x\vert$ at $x_1,x_2$. $\square$

Proof

2. The detailed derivation of $q^{*}(p)$in Eq. (19). The first derivative of $\pi (p,q)$ in Eq. (14):

$$\begin{aligned} \frac{\partial \pi (p,q)}{\partial q}=p-c-\frac{p}{2(1-\alpha )}\left(\frac{q-z}{\sqrt{v^2+(q-z)^2}}+1\right); \end{aligned}$$

(A1)

The second derivative of $\pi (p,q)$:

$$\begin{aligned} \frac{\partial ^2\pi (p,q)}{\partial q^2}=-\frac{p}{2(1-\alpha )}\frac{v^2}{(v^2+(q-z)^2)^{3/2}}\le 0. \end{aligned}$$

(A2)

This means $\pi (p,q)$ is concave in q. Thus, $\frac{\partial \pi (p,q)}{\partial q}=0$ gives the optimal $q^{*}(p)$ in Eq. (19). $\square$

Proof

3. The detailed derivation of $\pi (p,q^{*}(p))$ in Eq. (20). Define $U(p)=\frac{1}{1-(1-\alpha )(1-\frac{c}{p})}$, then $q^{*}(p)=z+\frac{v(U-2)}{2\sqrt{U-1}}$ and

$$\begin{aligned}&\pi (p,q^{*}(p))\\ =&(p-c)\left(z+\frac{v(U-2)}{2\sqrt{U-1}}\right)- \frac{p}{1-\alpha }\frac{\sqrt{v^2+\frac{v^2(U-2)^2}{4(U-1)}}+\frac{v(U-2)}{2\sqrt{U-1}}}{2}\\ =&(p-c)z+(p-c)\frac{v(U-2)}{2\sqrt{U-1}}-\frac{p}{1-\alpha }\frac{\frac{vU}{2\sqrt{U-1}}+\frac{v(U-2)}{2\sqrt{U-1}}}{2}\\ =&(p-c)z+(p-c)\frac{v(U-2)}{2\sqrt{U-1}}-\frac{pv(U-1)}{2(1-\alpha )\sqrt{U-1}}\\ =&(p-c)z+v\frac{(1-\alpha )(p-c)(U-2)-p(U-1)}{2(1-\alpha )\sqrt{U-1}}\\ =&(p-c)z+v\frac{(1-\alpha )(p-c)(U-1)-(1-\alpha )(p-c)-p(U-1)}{2(1-\alpha )\sqrt{U-1}}\\ =&(p-c)z-v\frac{(U-1)(p-(1-\alpha )(p-c))+(1-\alpha )(p-c)}{2(1-\alpha )\sqrt{U-1}}\\ =&(p-c)z-\frac{v}{2}\left[ \frac{\sqrt{\frac{(1-\alpha )(p-c)}{p-(1-\alpha )(p-c)}}(p-(1-\alpha )(p-c))}{1-\alpha }+\sqrt{\frac{p-(1-\alpha )(p-c)}{(1-\alpha )(p-c)}}(p-c)\right] \\ =&(p-c)z-\frac{v}{2}\left[ \sqrt{\frac{(p-c)(p-(1-\alpha )(p-c))}{1-\alpha }}+\sqrt{\frac{(p-(1-\alpha )(p-c))(p-c)}{1-\alpha }}\right] \\ =&(p-c)z-\frac{v}{\sqrt{1-\alpha }}\sqrt{(p-c)(p-(1-\alpha )(p-c))}\\ =&(p-c)z-\frac{v}{\sqrt{1-\alpha }}\sqrt{(p-c)(\alpha p+(1-\alpha )c)}. \end{aligned}$$

$\square$

Proof

4. The detailed derivation of Remark 1. $\pi (p,q^{*}(p))\ge 0$ means $p-c\ge \frac{\sigma ^2}{\mu ^2}\frac{1}{1-\alpha }(\alpha p+(1-\alpha )c)$, and then $(1-\frac{\sigma ^2}{\mu ^2}\frac{\alpha }{1-\alpha })p\ge (1+\frac{\sigma ^2}{\mu ^2})c$. So we have that $1-\frac{\sigma ^2}{\mu ^2}\frac{\alpha }{1-\alpha }>0$ and $p\ge \frac{(1+\frac{\sigma ^2}{\mu ^2})c}{1-\frac{\sigma ^2}{\mu ^2}\frac{\alpha }{1-\alpha }}$, they can be simplified as $\alpha <\frac{\mu ^2}{\mu ^2+\sigma ^2}$ and $p\ge (1+\frac{\sigma ^2}{\mu ^2-\alpha (\mu ^2+\sigma ^2)})c$. $\square$

Appendix B: Proofs

Proof

1. Proof of Theorem 1. Define $X(p)=\sqrt{\frac{\alpha p+(1-\alpha )c}{(1-\alpha )(p-c)}}\frac{\sigma }{\mu }-1$, we have that X(p) is decreasing in p and $X(p)\in (\sqrt{\frac{\alpha }{1-\alpha }}\frac{\sigma }{\mu }-1,0)$ for $(1+\frac{\sigma ^2}{\mu ^2-\alpha (\mu ^2+\sigma ^2)})c< p<\infty$, where $\sqrt{\frac{\alpha }{1-\alpha }}\frac{\sigma }{\mu }-1\in [-1,0)$ for $0\le \alpha <\frac{\mu ^2}{\mu ^2+\sigma ^2}$. The first derivative of X(p) is $\frac{\partial X(p)}{\partial p}=-\frac{1}{2}\frac{1}{p-c}\frac{c}{\alpha p+(1-\alpha )c}(X(p)+1)$. Using the new function X(p), we transfer $\pi (p,q^{*}(p))$ to

$$\begin{aligned} \pi (p,q^{*}(p))=-(p-c)X(p)z. \end{aligned}$$

(B3)

The first derivative of $\pi (p,q^{*}(p))$:

$$\begin{aligned} \begin{aligned} \frac{\partial \pi (p,q^{*}(p))}{\partial p}&=-(p-c)X(p)\frac{\partial z}{\partial p}-(X(p)+(p-c)\frac{\partial X(p)}{\partial p})z\\&=\left[ \eta (p)(1-\frac{c}{p})X(p)-\left( X(p)-\frac{1}{2}\frac{c}{\alpha p+(1-\alpha )c}(X(p)+1)\right) \right] z\\&\triangleq W(p)z, \end{aligned} \end{aligned}$$

(B4)

where $W(p)=\eta (p)(1-\frac{c}{p})X(p)-(X(p)-\frac{1}{2}\frac{c}{\alpha p+(1-\alpha )c}(X(p)+1))$.

The second derivative:

$$\begin{aligned} \frac{\partial ^2\pi (p,q^{*}(p))}{\partial p^2}=\frac{\partial W(p)}{\partial p}z+W(p)\frac{\partial z}{\partial p}. \end{aligned}$$

(B5)

In order to prove the unimodality of $\pi (p,q^{*}(p))$, we consider the sign of $\frac{\partial ^2\pi (p,q^{*}(p))}{\partial p^2}\mid _{\frac{\partial \pi (p,q^{*}(p))}{\partial p}=0}$, that is, the sign of $\frac{\partial W(p)}{\partial p}\mid _{W(p)=0}$. The condition $W(p)=0$ means the fact $\eta (p)=\frac{1}{2}\frac{p}{p-c}[2-\frac{c}{\alpha p+(1-\alpha )c}(1+\frac{1}{X(p)})]$. Then

$$\begin{aligned}&\frac{\partial W(p)}{\partial p}\mid _{W(p)=0}\\ =&\frac{\partial \eta (p)}{\partial p}(1-\frac{c}{p})X(p)+\eta (p)\left( \frac{c}{p^2}X(p)+\left( 1-\frac{c}{p}\right) \frac{\partial X(p)}{\partial p}\right) -\frac{\partial X(p)}{\partial p}\\&+\frac{1}{2}\left( -\frac{c\alpha }{(\alpha p+(1-\alpha )c)^2}(X(p)+1)+ \frac{c}{\alpha p+(1-\alpha )c}\frac{\partial X(p)}{\partial p}\right) \\ =&\frac{\partial \eta (p)}{\partial p}\left( 1-\frac{c}{p}\right) X(p)+\eta (p)\frac{c}{p^2}X(p)\\&+\left( \eta (p)\left( 1-\frac{c}{p}\right) -1+\frac{1}{2} \frac{c}{\alpha p+(1-\alpha )c}\right) \frac{\partial X(p)}{\partial p}\\&-\frac{1}{2}\frac{c\alpha }{(\alpha p+(1-\alpha )c)^2}(X(p)+1)\\ =&\frac{\partial \eta (p)}{\partial p} \left( 1-\frac{c}{p}\right) X(p)+\eta (p)\frac{c}{p^2}X(p)\\&+\left( 1-\frac{1}{2}\frac{c}{\alpha p+(1-\alpha )c} \left( 1+\frac{1}{X(p)}\right) -1 +\frac{1}{2}\frac{c}{\alpha p+(1-\alpha )c}\right) \frac{\partial X(p)}{\partial p}\\&-\frac{1}{2}\frac{c\alpha }{(\alpha p+(1-\alpha )c)^2}(X(p)+1)\\ =&\frac{\partial \eta (p)}{\partial p}(1-\frac{c}{p})X(p)+\eta (p)\frac{c}{p^2}X(p)-\frac{1}{2}\frac{c}{\alpha p+(1-\alpha )c}\frac{1}{X(p)}\frac{\partial X(p)}{\partial p}\\&-\frac{\alpha c}{2(\alpha p+(1-\alpha )c)^2}(X(p)+1). \end{aligned}$$

We have that $\frac{\partial W(p)}{\partial p}\mid _{W(p)=0}<0$ for $\frac{\partial \eta (p)}{\partial p}>0$, $-1<X(p)<0$, $\eta (p)>0$, $\frac{\partial X(p)}{\partial p}<0$. Therefore, $\pi (p,q^{*}(p))$ is unimodal in p, and $p^{*}$ is obtained by $\frac{\partial \pi (p,q^{*}(p))}{\partial p}=0$, that is $\eta (p)=\frac{1}{2}\frac{p}{p-c}[2-\frac{c}{\alpha p+(1-\alpha )c}(1+\frac{1}{X(p)})]\triangleq g_{\alpha }(p)$, where $g_{\alpha }(p)$ is defined in Eq. (23).

Define $\gamma (p)=2-\frac{c}{\alpha p+(1-\alpha )c}(1+\frac{1}{X(p)})$, then $g_{\alpha }(p)=\frac{1}{2}\frac{p}{p-c}\gamma (p)$. Since $-1<X(p)<0$, $1+\frac{1}{X(p)}<0$ and $X(p)+1>0$, we have that $\gamma (p)>0$ and

$$\begin{aligned} \begin{aligned} \frac{\partial \gamma (p)}{\partial p}&=\frac{c\alpha }{(\alpha p+(1-\alpha )c)^2} \left( 1+\frac{1}{X(p)}\right) +\frac{c}{\alpha p+(1-\alpha )c} \frac{1}{X(p)^2}\frac{\partial X(p)}{\partial p}\\&=\frac{c\alpha }{(\alpha p+(1-\alpha )c)^2} \left( 1+\frac{1}{X(p)}\right) -\frac{c}{\alpha p+(1-\alpha )c} \frac{1}{2}\frac{1}{p-c}\frac{c}{\alpha p+(1-\alpha )c}\frac{X(p)+1}{X(p)^2}\\&=\frac{c}{(\alpha p+(1-\alpha )c)^2}\left[ \alpha \left( 1+\frac{1}{X(p)}\right) -\frac{1}{2}\frac{c}{p-c}\frac{X(p)+1}{X(p)^2}\right] <0. \end{aligned} \end{aligned}$$

(B6)

$\frac{1}{2}\frac{p}{p-c}$ and $\gamma (p)$ are both positive and decreasing in p. Thus, $g_{\alpha }(p)$ is decreasing in p. Moreover, based on the fact that $\frac{\partial ^2 (f_{1}(p)f_{2}(p))}{\partial p^2}=\frac{\partial ^2 f_{1}(p)}{\partial p^2}f_{2}(p)+f_{1}(p)\frac{\partial ^2 f_{2}(p)}{\partial p^2}+2\frac{\partial f_{1}(p)}{\partial p}\frac{\partial f_{2}(p)}{\partial p}$, the second derivative of $g_{\alpha }(p)$ is $\frac{\partial ^2 g_{\alpha }(p)}{\partial p^2}=\frac{c}{(p-c)^3}\gamma (p)+\frac{p}{2(p-c)}\frac{\partial ^2 \gamma (p)}{\partial p^2}-\frac{c}{(p-c)^2}\frac{\partial \gamma (p)}{\partial p}$. We have that $\frac{\partial ^2 g_{\alpha }(p)}{\partial p^2}$ is positive, ie. $g_{\alpha }(p)$ is convex in p, for $\gamma (p)>0$, $\frac{\partial \gamma (p)}{\partial p}<0$, and $\frac{\partial ^2 \gamma (p)}{\partial p^2}=-2\frac{c\alpha }{(\alpha p+(1-\alpha )c)^3}[\alpha (1+\frac{1}{X(p)})-\frac{1}{2}\frac{c}{p-c}\frac{X(p)+1}{X(p)^2}]+\frac{c}{(\alpha p+(1-\alpha )c)^2}[-\frac{\alpha }{X(p)^2}\frac{\partial X(p)}{\partial p}+\frac{1}{2}\frac{c}{(p-c)^2}\frac{X(p)+1}{X(p)^2}+\frac{1}{2}\frac{c}{p-c}\frac{X(p)+2}{X(p)^3}\frac{\partial X(p)}{\partial p}]>0$. $\square$

Proof

2. Proof of Lemma 2. We first simplify $g_{\alpha }(p)$ using $h_{\alpha }(p)$, and then analyze the sign of $\frac{\partial h_{\alpha }(p)}{\partial \alpha }$ for the different p.

$$\begin{aligned} \begin{aligned} g_{\alpha }(p)&=\frac{1}{2}\frac{p}{p-c}\left[ 2-\frac{c}{\alpha p +(1-\alpha )c}\frac{\sqrt{\frac{\alpha p+(1-\alpha )c}{(1-\alpha )(p-c)}}\frac{\sigma }{\mu }}{\sqrt{\frac{\alpha p+(1-\alpha )c}{(1-\alpha )(p-c)}}\frac{\sigma }{\mu }-1}\right] \\&=\frac{1}{2}\frac{p}{p-c}\left[ 2-\frac{\sigma }{\mu }\frac{c}{p-c}\cdot \frac{1}{\frac{\sigma }{\mu } \left( \alpha +\frac{c}{p-c}\right) -\sqrt{(1-\alpha )\left( \alpha +\frac{c}{p-c}\right) }}\right] \\&\triangleq \frac{1}{2}\frac{p}{p-c}\left[ 2-\frac{\sigma }{\mu }\frac{c}{p-c}\cdot \frac{1}{h_{\alpha }(p)}\right] , \end{aligned} \end{aligned}$$

(B7)

where $h_{\alpha }(p)=\frac{\sigma }{\mu }(\alpha +\frac{c}{p-c})-\sqrt{(1-\alpha )(\alpha +\frac{c}{p-c})}$. For a fixed p, the monotonicity of $g_{\alpha }(p)$ in $\alpha$ is related to the monotonicity of $h_{\alpha }(p)$ in $\alpha$.

The first derivative of $h_{\alpha }(p)$ with respect to $\alpha$:

$$\begin{aligned} \frac{\partial h_{\alpha }(p)}{\partial \alpha }=\frac{\sigma }{\mu }-\frac{(1-\alpha )-\left( \alpha +\frac{c}{p-c}\right) }{2\sqrt{(1-\alpha )\left( \alpha +\frac{c}{p-c}\right) }}; \end{aligned}$$

(B8)

The second derivative of $h_{\alpha }(p)$ with respect to $\alpha$:

$$\begin{aligned} \frac{\partial ^2 h_{\alpha }(p)}{\partial \alpha ^2}=\frac{1}{4}\frac{\left( \frac{p}{p-c}\right) ^2}{\left[ (1-\alpha )\left( \alpha +\frac{c}{p-c}\right) \right] ^{3/2}}>0. \end{aligned}$$

(B9)

This means $h_{\alpha }(p)$ is convex in $\alpha$. Next we will discuss the sign of $\frac{\partial h_{\alpha }(p)}{\partial \alpha }$.

Define $l(\alpha ,p)$ as $l(\alpha ,p)=\frac{\partial h_{\alpha }(p)}{\partial \alpha }=\frac{\sigma }{\mu }-\frac{1}{2}(\sqrt{\frac{1-\alpha }{\alpha +\frac{c}{p-c}}}-\sqrt{\frac{\alpha +\frac{c}{p-c}}{1-\alpha }})$ is a function of $\alpha$ and p, it is increasing in $\alpha$ and decreasing in p, next we will derive the ranges of $\alpha$ and p which satisfy $l(\alpha ,p)>0$ and $l(\alpha ,p)<0$ respectively. For a given $\alpha$, $l(\alpha ,p)$ is decreasing to the lower bound $l(\alpha ,\infty )=\frac{\sigma }{\mu }-\frac{(1-\alpha )-\alpha }{2\sqrt{(1-\alpha )\alpha }}$ when p increases to $\infty$. We first analyze the sign of the lower bound $l(\alpha ,\infty )$ and then the sign of $l(\alpha ,p)$. The lower bound is increasing in $\alpha$, and it is negative when $0\le \alpha <\frac{\delta }{1+\delta }$ and positive when $\frac{\delta }{1+\delta }<\alpha <\alpha _{max}$, where $\frac{\delta }{1+\delta }$ is the solution of $l(\alpha ,\infty )=0$. Thus, the sign of $l(\alpha ,p)$ is divided into the following two cases.

For $\forall \alpha \in [0,\frac{\delta }{1+\delta })$, the lower bound $l(\alpha ,\infty )<0$, and because $l(\alpha ,p)$ decreases in p, there exists a unique solution $p_0$, which is defined as $p_0=(1+\frac{1}{(1-\alpha )\delta -\alpha })c$, satisfying $l(\alpha ,p_0)=0$ (We know $p_0$ is indeed in the range $[p_{min},\infty )$ for the fact that $l(\alpha ,p_{min})=\frac{1}{2}(\frac{\sigma }{\mu }+\frac{\mu }{\sigma })>0$, where $p_{min}=(1+\frac{\sigma ^2}{\mu ^2-\alpha (\mu ^2+\sigma ^2)})c$). For $p> p_{0}$, we have that $l(\alpha ,p)<0$ and $g_{\alpha }(p)$ decreases in $\alpha$; for $p\le p_{0}$, we have that $l(\alpha ,p)\ge 0$ and $g_{\alpha }(p)$ increases in $\alpha$.

For $\forall \alpha \in [\frac{\delta }{1+\delta },\alpha _{max})$, the lower bound $l(\alpha ,\infty )\ge 0$, and because $l(\alpha ,p)$ decreases in p, the first order condition $l(\alpha ,p)=0$ has no solution. In this case, $l(\alpha ,p)>l(\alpha ,\infty )\ge 0$ and $g_{\alpha }(p)$ is increasing in $\alpha$ for all $p\in [(1+\frac{\sigma ^2}{\mu ^2-\alpha (\mu ^2+\sigma ^2)})c,\infty )$. $\square$

Proof

3. Proof of Theorem 3. The outline of the proof is the following:

Part 1: We analyze the sensitivity of $p^{*}(\alpha )$. Based on the results in Lemma 2, the analysis is divided into two ranges: $[0,\frac{\delta }{1+\delta })$ and $[\frac{\delta }{1+\delta },\alpha _{max})$. For $\alpha \in [\frac{\delta }{1+\delta },\alpha _{max})$, $g_{\alpha }(p)$ is increasing in $\alpha$ for all p, thus, $p^{*}(\alpha )$ is increasing in $\alpha$. For $\alpha \in [0,\frac{\delta }{1+\delta })$, $g_{\alpha }(p)$ is increasing in $\alpha$ for $p\le p_{0}(\alpha )$ and decreasing in $\alpha$ for $p>p_{0}(\alpha )$, we need to compare $p^{*}(\alpha )$ and $p_0(\alpha )$, which is equivalent to compare $g_{\alpha }(p_0(\alpha ))$ and $\eta (p_0(\alpha ))$. For the $\alpha$ satisfying that $g_{\alpha }(p_0(\alpha ))<\eta (p_0(\alpha ))$, it satisfies that $p^{*}(\alpha )<p_0(\alpha )$ and $p^{*}(\alpha )$ is increasing, and vice versa.

Part 2: We analyze the sensitivity of $q^{*}(\alpha )$ according to Eq. (24). To analyze the monotonicity of $(1-\alpha )(1-\frac{c}{p^{*}})$, we introduce a variable $G^{*}(\alpha )$ which is a function of $\alpha$ and $p^{*}(\alpha )$, and we rewrite $q^{*}(\alpha )$ as the function of $y(p^{*})$ and $G^{*}(\alpha )$, then analyze the monotonicity of $G^{*}$ as $\alpha$ increases.

Part 1: The sensitivity of $p^{*}(\alpha )$.

First, we analyze the sensitivity of $p^{*}(\alpha )$, which is the unique intersection of decreasing function $g_{\alpha }(p)$ and increasing function $\eta (p)$.

The monotonicity of $g_{\alpha }(p)$ in $\alpha \in [0,\alpha _{max})$ is divided into two ranges: $[0,\frac{\delta }{1+\delta })$ and $[\frac{\delta }{1+\delta },\alpha _{max})$.

For $\alpha \in [\frac{\delta }{1+\delta },\alpha _{max})$, $g_{\alpha }(p)$ is increasing in $\alpha$ for all p and then $p^{*}(\alpha )$ is increasing in $\alpha$, which is according to Theorem 1 and Lemma 2.

For $\alpha \in [0,\frac{\delta }{1+\delta })$, $g_{\alpha }(p)$ is increasing in $\alpha$ for $p\le p_{0}(\alpha )$ and decreasing in $\alpha$ for $p>p_{0}(\alpha )$, according to Lemma 2. In order to analyze the sensitivity of $p^{*}(\alpha )$, we need to compare $p^{*}(\alpha )$ and $p_0(\alpha )$. If $p^{*}(\alpha )\le p_0(\alpha )$, then $g_{\alpha }(p)$ is increasing in $\alpha$ at all points in range $p\in (p_{min},p_0(\alpha ))$ including $p=p^{*}(\alpha )$, which implies $p^{*}(\alpha )$ is increasing because it is the unique intersection of decreasing function $g_{\alpha }(p)$ and increasing function $\eta (p)$. Similarly, if $p^{*}(\alpha )>p_0(\alpha )$, then $g_{\alpha }(p)$ is decreasing in $\alpha$ at all points in range $p\in (p_0(\alpha ),\infty )$ including $p=p^{*}(\alpha )$, which implies $p^{*}(\alpha )$ is decreasing in $\alpha$.

To compare $p^{*}(\alpha )$ and $p_0(\alpha )$, we need to analyze the solutions of $p^{*}(\alpha )=p_0(\alpha )$, which is equivalent to $g_{\alpha }(p_0(\alpha ))=\eta (p_0(\alpha ))$, i.e., $g_{\alpha }((1+\frac{1}{(1-\alpha )\delta -\alpha })c)=\eta ((1+\frac{1}{(1-\alpha )\delta -\alpha })c)$. Specifically, $g_{\alpha }((1+\frac{1}{(1-\alpha )\delta -\alpha })c)=k_1 \alpha +k_2$, where $k_1=-\frac{(1+\delta )^2}{2\delta }<0$ and $k_2=\frac{3+\delta }{2}>0$ are constants. Therefore, $g_{\alpha }((1+\frac{1}{(1-\alpha )\delta -\alpha })c)$ decreases from $k_2$ to $k_1\frac{\delta }{1+\delta }+k_2$ when $\alpha$ increases from 0 to $\frac{\delta }{1+\delta }$. Note that $\eta ((1+\frac{1}{(1-\alpha )\delta -\alpha })c)$ is increasing in $\alpha$ from $\eta (c+\frac{c}{\delta })$ to $\infty$ when $\alpha$ increases from 0 to $\frac{\delta }{1+\delta }$.

If $\eta (c+\frac{c}{\delta })\ge k_2$, then $g_{\alpha }(p_0(\alpha ))\le \eta (p_0(\alpha ))$ for all $\alpha \in [0,\frac{\delta }{1+\delta })$. Note $p=p^{*}(\alpha )$ is the value which satisfies $g_{\alpha }(p)=\eta (p)$. Therefore, we have that $p^{*}(\alpha )\le p_0(\alpha )$ and then $p^{*}(\alpha )$ is increasing in $\alpha$ for all $\alpha \in [0,\frac{\delta }{1+\delta })$.

If $\eta (c+\frac{c}{\delta })< k_2$, then there exists a unique $\alpha ^{*}\in [0,\frac{\delta }{1+\delta })$ which is the solution of

$$\begin{aligned} k_1 \alpha +k_2=\eta \left( \left( 1+\frac{1}{(1-\alpha )\delta -\alpha }\right) c\right) . \end{aligned}$$

(B10)

For $0\le \alpha <\alpha ^{*}$, we have that $g_{\alpha }(p_0(\alpha ))>\eta (p_0(\alpha ))$, which implies $p^{*}(\alpha )> p_0(\alpha )$ and then $p^{*}(\alpha )$ is decreasing in $\alpha$. Similarly, for $\alpha ^{*}\le \alpha <\frac{\delta }{1+\delta }$, we have that $p^{*}(\alpha )$ is increasing in $\alpha$.

Part 2: The sensitivity of $q^{*}(\alpha )$.

Define $G=(1-\alpha )(1-\frac{c}{p})$, and then $p=\frac{c}{1-\frac{G}{1-\alpha }}$, $\frac{p}{p-c}=\frac{1-\alpha }{G}$, $\alpha +(1-\alpha )\frac{c}{p}=1-G$ and $\alpha p+(1-\alpha )c=\frac{c}{1-\frac{G}{1-\alpha }}(1-G)$. Since $p^{*}$ is the solution of $g_{\alpha }(p)=\eta (p)$, so $G^{*}=(1-\alpha )(1-\frac{c}{p^{*}})$ is the solution of $g_{\alpha }(\frac{c}{1-\frac{G}{1-\alpha }})=\eta (\frac{c}{1-\frac{G}{1-\alpha }})$. Define $\hat{g}_{\alpha }(G)=g_{\alpha }(\frac{c}{1-\frac{G}{1-\alpha }})$ and $\hat{\eta }_{\alpha }(G)=\eta (\frac{c}{1-\frac{G}{1-\alpha }})$. Specifically,

$$\begin{aligned} \begin{aligned} \hat{g}_{\alpha }(G)&=g_{\alpha }\left( \frac{c}{1-\frac{G}{1-\alpha }}\right) =\frac{1-\alpha }{2G}\left[ 2-\frac{1-\frac{G}{1-\alpha }}{c}\frac{c}{1-G} \left( 1+\frac{1}{\sqrt{\frac{1-G}{G}}\frac{\sigma }{\mu }-1}\right) \right] \\&=\frac{1-\alpha }{G}-\frac{1}{2} \left( \frac{1-\alpha }{G}-1\right) \frac{1}{1-G}\left( 1+\frac{1}{\sqrt{\frac{1-G}{G}}\frac{\sigma }{\mu }-1}\right) \\&=\frac{1}{G}\left[ 1-\frac{1}{2}\frac{1}{1-G} \left( 1+\frac{1}{\sqrt{\frac{1-G}{G}}\frac{\sigma }{\mu }-1}\right) \right] (1-\alpha )+\frac{1}{2}\frac{1}{1-G}\left( 1+\frac{1}{\sqrt{\frac{1-G}{G}}\frac{\sigma }{\mu }-1}\right) . \end{aligned} \end{aligned}$$

(B11)

Note that $\frac{\partial \hat{g}_{\alpha }(G)}{\partial G}=\frac{\partial g(p)}{\partial p}\frac{\partial p}{\partial G}<0$, $\hat{g}_{\alpha }(G)$ is a decreasing function of G. And for a fixed G, it is a decreasing function of $\alpha$, for $1+\frac{1}{\sqrt{\frac{1-G}{G}}\frac{\sigma }{\mu }-1}=1+\frac{1}{X(p)}<0$, where X(p) is defined in the proof of Theorem 1. Since $\hat{\eta }_{\alpha }(G)=\eta (\frac{c}{1-\frac{G}{1-\alpha }})$, thus, $\hat{\eta }_{\alpha }(G)$ is an increasing function of G, and for a fixed G, it is increasing in $\alpha$.

In conclusion, when $\alpha$ increases, we have that $\hat{g}_{\alpha }(G)$ decreases for all G and $\hat{\eta }_{\alpha }(G)$ increases for all G, therefore $G^{*}$, the solution of $\hat{g}_{\alpha }(G)=\hat{\eta }_{\alpha }(G)$, is decreasing in $\alpha$ (see Fig. 5).

If $p^{*}(\alpha )$ increases in $\alpha$, then

$$\begin{aligned} q^{*}(\alpha )=y(p^{*})\mu +\frac{y(p^{*})\sigma (\frac{1}{1-G^{*}}-2)}{2\sqrt{\frac{1}{1-G^{*}}-1}} \end{aligned}$$

(B12)

decreases in $\alpha$, for y(p) decreases in p, $\frac{\frac{1}{1-G^{*}}-2}{\sqrt{\frac{1}{1-G^{*}}-1}}$ increases in $G^{*}$ and $G^{*}$ decreases in $\alpha$. $\square$

Fig. 5

The sensitivity of $G^{*}$ with respect to confidence level $\alpha$

Proof

4. Proof of Example 1. According to Theorem 3, there exists a unique $\alpha ^{*}\in [0,\frac{\delta }{1+\delta })$ which is the solution of

$$\begin{aligned} k_1 \alpha +k_2=bc(1+\frac{1}{(1-\alpha )\delta -\alpha }). \end{aligned}$$

(B13)

Equation (B13) can be written as a quadratic equation with respect to $\alpha$. Its roots are $\frac{\pm \sqrt{(1+b^{2}c^2)\delta ^2+2bc\delta }+(\delta +2-bc)\delta }{(1+\delta )^2}$, the existence and uniqueness of $\alpha ^{*}\in [0,\frac{\delta }{1+\delta })$ which is proved in Theorem 3 guarantees the value in the square root is positive. Moreover, the larger root satisfies

$$\begin{aligned} \frac{\sqrt{(1+b^{2}c^2)\delta ^2+2bc\delta }+(\delta +2-bc)\delta }{(1+\delta )^2}>\frac{\delta }{1+\delta } \end{aligned}$$

(B14)

such that it is infeasible. Therefore, $\alpha ^{*}=\frac{-\sqrt{(1+b^{2}c^2)\delta ^2+2bc\delta }+(\delta +2-bc)\delta }{(1+\delta )^2}$. $\square$

Proof

5. Proof of Theorem 4. Define $B(p)=\sigma \sqrt{\frac{\alpha p+(1-\alpha )c}{(1-\alpha )(p-c)}}$, B(p) is decreasing in p and $\frac{\partial B(p)}{\partial p}=-\frac{1}{2}\frac{1}{p-c}\frac{c}{\alpha p+(1-\alpha )c}B(p)$. Using the new function B(p), we transfer $\pi (p,q^{*}(p))$ to

$$\begin{aligned} \pi (p,q^{*}(p))=(p-c)y(p)-(p-c)B(p). \end{aligned}$$

(B15)

The first derivative of $\pi (p,q^{*}(p))$:

$$\begin{aligned} \frac{\partial \pi (p,q^{*}(p))}{\partial p}=\left[ 1-\left( 1-\frac{c}{p}\right) \eta (p)\right] y(p)-\left( 1-\frac{1}{2}\frac{c}{\alpha p+(1-\alpha )c}\right) B(p). \end{aligned}$$

(B16)

Based the fact that $y(p)\ge B(p)$ and $1-\frac{1}{2}\frac{c}{\alpha p+(1-\alpha )c}\ge 0$, we assume that $0\le 1-(1-\frac{c}{p})\eta (p) \le 1-\frac{1}{2}\frac{c}{\alpha p+(1-\alpha )c}$, i.e.,

$$\begin{aligned} \frac{1}{2}\frac{c}{\alpha p+(1-\alpha )c}\frac{p}{p-c}\le \eta (p)\le \frac{p}{p-c} \end{aligned}$$

(B17)

to ensure there exists a solution of $\frac{\partial \pi (p,q^{*}(p))}{\partial p}=0$. It implies that $p\in [p_1, p_2]$, where $p_1$ is the unique solution of $\eta (p)=\frac{1}{2}\frac{c}{\alpha p+(1-\alpha )c}\frac{p}{p-c}$ and $p_2$ is the unique solution of $\eta (p)=\frac{p}{p-c}$. For $p<p_1$, $\frac{\partial \pi (p,q^{*}(p))}{\partial p}>0$ and $\pi (p,q^{*}(p))$ is strictly increasing in p; for $p>p_2$, $\frac{\partial \pi (p,q^{*}(p))}{\partial p}<0$ and $\pi (p,q^{*}(p))$ is strictly decreasing in p. Therefore, we only need to prove the unimodality property of $\pi (p,q^{*}(p))$ for $p\in [p_1, p_2]$, and the unique maximizer will be the global optimum solution.

The second derivative:

$$\begin{aligned} \begin{aligned} \frac{\partial ^2\pi (p,q^{*}(p))}{\partial p^2} =&\left[ \frac{1}{p}\left( 1-\frac{c}{p}\right) \eta ^{2}(p) -\frac{1}{p}\left( \frac{c}{p}+1\right) \eta (p)-\left( 1-\frac{c}{p}\right) \frac{\partial \eta (p)}{\partial p}\right] y(p)\\&+\frac{c^2}{4(p-c)(\alpha p+(1-\alpha )c)^2}B(p). \end{aligned} \end{aligned}$$

(B18)

In order to prove the unimodality of $\pi (p,q^{*}(p))$, we consider the sign of $\frac{\partial ^2\pi (p,q^{*}(p))}{\partial p^2}\mid _{\frac{\partial \pi (p,q^{*}(p))}{\partial p}=0}$. The condition $\frac{\partial \pi (p,q^{*}(p))}{\partial p}=0$ means the fact $B(p)=[1-(1-\frac{c}{p})\eta (p)]\frac{\alpha p+(1-\alpha )c}{\alpha p+(1-\alpha )c-\frac{c}{2}}y(p)$. Then

$$\begin{aligned} \begin{aligned}&\frac{\partial ^2\pi (p,q^{*}(p))}{\partial p^2}\mid _{\frac{\partial \pi (p,q^{*}(p))}{\partial p}=0}\\ \le&\frac{1}{p}\left[ \left( 1-\frac{c}{p}\right) \eta (p)-1\right] \eta (p)y(p)\\&+\frac{c^2}{4(p-c)(\alpha p+(1-\alpha )c)^2}\left[ 1-\left( 1-\frac{c}{p}\right) \eta (p)\right] \frac{\alpha p+(1-\alpha )c}{\alpha p+(1-\alpha )c-\frac{c}{2}}y(p)\\ =&\left[ 1-\left( 1-\frac{c}{p}\right) \eta (p)\right] y(p) \left[ -\frac{1}{p}\eta (p)+\frac{c^2}{4(p-c)(\alpha p+(1-\alpha )c)\left( \alpha p+(1-\alpha )c-\frac{c}{2}\right) }\right] \\ \le&\left[ 1-\left( 1-\frac{c}{p}\right) \eta (p)\right] y(p) \left[ -\frac{1}{p}\frac{1}{2}\frac{c}{\alpha p+(1-\alpha )c}\frac{p}{p-c}\right. \\&\left. +\frac{c^2}{4(p-c)(\alpha p+(1-\alpha )c)(\alpha p+(1-\alpha )c-\frac{c}{2})}\right] \\ =&\left[ 1-\left( 1-\frac{c}{p}\right) \eta (p)\right] y(p)\frac{-c\alpha }{2(\alpha p+(1-\alpha )c)\left( \alpha p+(1-\alpha )c-\frac{c}{2}\right) }\\ \le&0. \end{aligned} \end{aligned}$$

The unimodality property of $\pi (p,q^{*}(p))$ is proved and $p^{*}$ satisfies $\frac{\partial \pi (p,q^{*}(p))}{\partial p}=0$, i.e., $[1-(1-\frac{c}{p})\eta (p)]y(p)-(1-\frac{1}{2}\frac{c}{\alpha p+(1-\alpha )c})B(p)=0$. And then the optimal order quantity is $q^{*}=y(p^{*})+\frac{\sigma (\frac{1}{1-(1-\alpha )(1-c/p^{*})}-2)}{2\sqrt{\frac{1}{1-(1-\alpha )(1-c/p^{*})}-1}}$ following Eq. (19).

Next, we will analyze how the optimal $p^*$ changes with $\alpha$, where $p^*$ and $\alpha$ satisfy $\frac{\partial \pi (p,q^*(p))}{\partial p}=0$. Take the first derivative of the first-order condition $\frac{\partial \pi (p,q^*(p))}{\partial p}=0$ with respect to $\alpha$, we obtain

$$\begin{aligned} \frac{\partial ^2\pi (p,q^{*}(p))}{\partial p^2}\frac{dp}{d\alpha }-\frac{2p-c}{4(\alpha p+(1-\alpha )c)}B(P)=0. \end{aligned}$$

Since when $p=p^*$, $\frac{\partial ^2\pi (p,q^{*}(p))}{\partial p^2}\le 0$, and $\frac{p(p-c)}{2(\alpha p+(1-\alpha )c)^2}B(P)\ge 0$, then we obtain $\frac{dp^*}{d\alpha }\le 0$. $\square$